A synthetic mixture of nitrogen and Argon has a density of `1.4 g L^(-1)` at `0^@C`. Calculate the average molecular weight. Find out the volume percentage of nitrogen in the mixture.

Explanation

Molecular weight (M) can be obtained from density DRT (d) as,
`M = (dRT)/(P)` Average molecular weight of the mixture
`= (1.4 xx 0.0821 xx 273)/(1) = 31.4`
If the % volume of `N_2` is .x.
`3.14 = (x xx 28 + (100 - x)40)/(100) implies 12x = 840 " or " x = 70`
The volume percentage of `N_2` in the mixture = 70